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Question: A bag contains 3 white, 3 black and 2 red balls. One by one three balls are drawn without replacing ...

A bag contains 3 white, 3 black and 2 red balls. One by one three balls are drawn without replacing them. The probability that the third ball is red, is

A

12\frac { 1 } { 2 }

B

13\frac { 1 } { 3 }

C

23\frac { 2 } { 3 }

D

14\frac { 1 } { 4 }

Answer

14\frac { 1 } { 4 }

Explanation

Solution

Let R stand for drawing red ball for drawing black ball and for drawing white ball.

Then required probability

=P(WWR)+P(BBR)+P(WBR)+P(BWR)+P(WRR)+= P ( W W R ) + P ( B B R ) + P ( W B R ) + P ( B W R ) + P ( W R R ) +

P(BRR)+P(RWR)+P(RBR)P ( B R R ) + P ( R W R ) + P ( R B R )

=3⋅2⋅28⋅7⋅6+3⋅2⋅28⋅7⋅6+3⋅3⋅28⋅7⋅6+3⋅3⋅28⋅7⋅6+3⋅2⋅18⋅7⋅6= \frac { 3 \cdot 2 \cdot 2 } { 8 \cdot 7 \cdot 6 } + \frac { 3 \cdot 2 \cdot 2 } { 8 \cdot 7 \cdot 6 } + \frac { 3 \cdot 3 \cdot 2 } { 8 \cdot 7 \cdot 6 } + \frac { 3 \cdot 3 \cdot 2 } { 8 \cdot 7 \cdot 6 } + \frac { 3 \cdot 2 \cdot 1 } { 8 \cdot 7 \cdot 6 }

=256+256+356+356+156+156+156+156=14= \frac { 2 } { 56 } + \frac { 2 } { 56 } + \frac { 3 } { 56 } + \frac { 3 } { 56 } + \frac { 1 } { 56 } + \frac { 1 } { 56 } + \frac { 1 } { 56 } + \frac { 1 } { 56 } = \frac { 1 } { 4 }.